Question

2sin(2x+pi/6)+1=корень3×sin2x+cosx

Answer 1

2Sin(2x + π/6) + 1 = √3Sin2x + Cosx

2(Sin2xCosπ/6 + Cos2xSinπ/6) + 1 = √3Sin2x + Cosx

2(Sin2x * √3/2 + Cos2x * 1/2) + 1 = √3Sin2x + Cosx

√3Sin2x + Cos2x + 1 = √3Sin2x + Cosx

Cos2x - Cosx + 1 = 0

2Cos²x - 1 - Cosx + 1 = 0

2Cos²x - Cosx = 0

Cosx(2Cosx - 1) = 0

1) Cosx = 0

x = π/2 + πn , n ∈ Z

2) 2Cosx - 1 = 0

2Cosx = 1

Cosx = 1/2

x= ± arcCos1/2 + 2πn , n ∈ Z

x = ± π/3 + 2πn , n ∈ Z

Answer 2

To solve the equation 2sin(2x + π/6) + 1 = √3*sin(2x) + cos(x), we can follow these steps:

Step 1: Simplify the equation by expanding the trigonometric functions and distributing the terms:

2sin(2x)cos(π/6) + 2cos(2x)sin(π/6) + 1 = √3*sin(2x) + cos(x)

Since cos(π/6) = √3/2 and sin(π/6) = 1/2, the equation becomes:

√3sin(2x) + √3cos(2x)/2 + 1 = √3*sin(2x) + cos(x)

Step 2: Move the terms involving sin(2x) to one side of the equation and the remaining terms to the other side:

√3*cos(2x)/2 + 1 = cos(x)

Subtract cos(x) from both sides:

√3*cos(2x)/2 - cos(x) + 1 = 0

Step 3: Combine the terms involving cos(x) by finding a common denominator:

√3cos(2x)/2 - 2cos(x)/2 + 2/2 = 0

(√3cos(2x) - 2cos(x) + 2)/2 = 0

Step 4: Multiply the entire equation by 2 to eliminate the fraction:

√3cos(2x) - 2cos(x) + 2 = 0

Step 5: Now we have a quadratic-like equation in terms of cos(x). Let's substitute u = cos(x):

√3cos(2x) - 2u + 2 = 0

Step 6: Rewrite the equation in terms of u:

√3*(2u^2 - 1) - 2u + 2 = 0

Step 7: Expand and rearrange the equation:

2√3u^2 - √3 - 2u + 2 = 0

2√3u^2 - 2u - √3 + 2 = 0

Step 8: Solve this quadratic equation. Unfortunately, the solutions are not easy to express in simple terms and require the use of numerical methods or calculators.

Note: It's possible that there might be a mistake in the equation or the equation might not have nice, simple solutions. Please double-check the equation or provide any additional information if needed.